1. In quadrilateral ACBD, AC = AD and AB bisects \(\angle{A}\) (see figure). Show that \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ABD}\). What can you say about BC and BD ?

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In \(\triangle{ABC}\) and \(\triangle{ABD}\), we have,

AC = AD ....(Given)

As, AB bisects \(\angle{A}\),

Therefore, \(\angle{CAB}\) = \(\angle{DAB}\)

Also, AB is a common side,

So, we can say,

\(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ABD}\) ...(by SAS test of congruency)

Also, BC = BD ...(By CPCT)

Hence, proved.

2. ABCD is a quadrilateral in which AD = BC and \(\angle{DAB}\) = \(\angle{CBA}\) (see figure). Prove
that

(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{BAC}\)

(ii) BD = AC

(iii) \(\angle{ABD}\) = \(\angle{BAC}\)

(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{BAC}\)

(ii) BD = AC

(iii) \(\angle{ABD}\) = \(\angle{BAC}\)

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In \(\triangle{ABC}\) and \(\triangle{BAC}\), we have

AD = BC ...(Given)

Also,
\(\angle{DAB}\) = \(\angle{CBA}\) ...(Given)

And AB is a common side.

Therefore,
\(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{BAC}\) ...(SAS congruency test)

Hence, BD = AC ...(By CPCT)

and \(\angle{ABD}\) = \(\angle{BAC}\) ...(By CPCT)

Hence, proved.

3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

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In \(\triangle{AOD}\) and \(\triangle{BOC}\), we have,

Now, \(\angle{AOD}\) = \(\angle{BOC}\) ...(vertically opposite angles)

Also, \(\angle{DAO}\) = \(\angle{CBO}\) = \(90^\circ\)

and BD = BC ...(shown in the figure)

Therefore, \(\triangle{AOD}\) \(\displaystyle \cong \) \(\triangle{BOC}\) ...(SAS congruency test)

Hence, OA = OB ...(By CPCT)

Thus, we can say that, O is the mid-point of AB.

So, CD bisects AB.

Hence, proved

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{CDA}\).

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From figure, we have,

\(\angle{1}\) = \(\angle{2}\) (Vertically opposite angles) ...(i)

\(\angle{1}\) = \(\angle{6}\) (Corresponding angles) ...(ii)

\(\angle{6}\) = \(\angle{4}\) (Corresponding angles) ...(iii)

From Equations, (i), (ii) and (iii), we have

\(\angle{1}\) = \(\angle{4}\) and \(\angle{2}\) = \(\angle{6}\) ...(iv)

In \(\triangle{ABC}\) and \(\triangle{CDA}\), we have

\(\angle{4}\) = \(\angle{2}\) ...(from (iii) and (iv))

\(\angle{5}\) = \(\angle{3}\) ...(Alternate angles)

and AC is a common side.

Therefore, we get,

\(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{CDA}\) ...(By SAS congruency test)

Hence, proved.

5. Line l is the bisector of a \(\angle{A}\) and \(\angle{B}\) is any point on l. BP and BQ are
perpendiculars from B to the arms of \(\angle{A}\) (see figure). show that

(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\)

(ii) BP = BQ or B is equidistant from the arms of \(\angle{A}\).

(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\)

(ii) BP = BQ or B is equidistant from the arms of \(\angle{A}\).

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In \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\), we have

\(\angle{APB}\) = \(\angle{AQB}\) = \(90^\circ\)

\(\angle{PAB}\) = \(\angle{QAB}\) ...(AB bisects \(\angle{PAQ}\))

AB is the common side.

Therefore, \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\) ...(By AAS congruency test)

Also, BP = BQ ...(By CPCT)

Thus, we can say that, B is equidistant from the arms of \(\angle{A}\).

Hence, proved.

6. In figure, AC = AE, AB = AD and \(\angle{BAD}\) = \(\angle{EAC}\). Show that BC = DE.

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In \(\triangle{ABC}\) and \(\triangle{ADE}\) , we have,

AB = AD ...(Given)

\(\angle{BAD}\) = \(\angle{EAC}\) ...(i)(Given)

On adding, \(\angle{DAC}\) on both sides in Eq. (i), we get,

\(\angle{BAD}\) + \(\angle{DAC}\) = \(\angle{EAC}\) + \(\angle{DAC}\)

i.e., \(\angle{BAC}\) = \(\angle{DAE}\)

and also, AC = AE ...(Given)

Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ADE}\) ...(By AAS congruency test)

Thus, BC = DE ...(By CPCT)

Hence, proved.

7. AB is a line segment and P is its mid-point. D and E are points on the same side of
AB such that \(\angle{BAD}\) = \(\angle{ABE}\) and \(\angle{EPA}\) = \(\angle{DPB}\)(see figure). Show that

(i) \(\triangle{DAP}\) \(\displaystyle \cong \) \(\triangle{EBP}\)

(ii) AD = BE

(i) \(\triangle{DAP}\) \(\displaystyle \cong \) \(\triangle{EBP}\)

(ii) AD = BE

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We have,

AP = BP ...(i)(Since, P is the mid-point of AB)

\(\angle{EPA}\) = \(\angle{DPB}\) ...(ii)(Given)

\(\angle{BAD}\) = \(\angle{ABE}\) ...(iii) (Given)

On adding, \(\angle{EPD}\) on both sides in Equation (ii),

we have,

\(\angle{EPA}\) + \(\angle{EPD}\) = \(\angle{DPB}\) + \(\angle{EPD}\)

\(\angle{DPA}\) = \(\angle{EPB}\) ...(iv)

Now, In \(\triangle{DAP}\) and \(\triangle{EBP}\),

We have,
\(\angle{DPA}\) = \(\angle{EPB}\) ...(from (iv)),

\(\angle{DAP}\) = \(\angle{EBP}\) ...(Given)

and AP = BP ...(From Eq. (i))

Therefore, \(\triangle{DAP}\) \(\displaystyle \cong \) \(\triangle{EBP}\) ...(By ASA congruency test)

Thus, AD = BE ...(By CPCT)

Hence, proved.

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is
joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure).

Show that

(i) \(\triangle{AMC}\) \(\displaystyle \cong \) \(\triangle{BMD}\)

(ii) \(\angle{DBC}\) is a right angle

(iii) \(\triangle{DBC}\) \(\displaystyle \cong \) \(\triangle{ACB}\)

(iv)CM = (1/2) AB

Show that

(i) \(\triangle{AMC}\) \(\displaystyle \cong \) \(\triangle{BMD}\)

(ii) \(\angle{DBC}\) is a right angle

(iii) \(\triangle{DBC}\) \(\displaystyle \cong \) \(\triangle{ACB}\)

(iv)CM = (1/2) AB

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Given :\(\angle{ACB}\) in which \(\angle{C}\) = \(90^\circ\) and M is the mid-point of AB.

To prove :

(i) \(\triangle{AMC}\) \(\displaystyle \cong \) \(\triangle{BMD}\)

(ii) \(\angle{DBC}\) is a right angle

(iii) \(\triangle{DBC}\) \(\displaystyle \cong \) \(\triangle{ACB}\)

(iv)CM = (1/2) AB

Construction : Produce CM to D, such that CM = MD. Join DB.

Proof : In \(\triangle{AMC}\) and \(\triangle{BMD}\), we have

AM = BM ...(M is the mid-point of AB)

CM = DM ...(Given)

and \(\angle{AMC}\) = \(\angle{BMD}\) ...(Vertically opposite angles)

Therefore, \(\triangle{AMC}\) \(\displaystyle \cong \) \(\triangle{BMD}\)(By SAS congruency test)

Hence, part(i) is proved.

Also, AC = DB ...(By CPCT)

\(\angle{1}\) = \(\angle{2}\) ...(Alternate angles) and (by CPCT)

Therefore, BD || CA and BC is transversal.

\(\angle{ACB}\) + \(\angle{DBC}\) = \(180^\circ\)

But, \(\angle{ACB}\) = \(90^\circ\) ...(given)

i.e., \(90^\circ\) + \(\angle{DBC}\) = \(180^\circ\)

i.e., \(\angle{DBC}\) = \(180^\circ\) - \(90^\circ\)

i.e., \(\angle{DBC}\) = \(90^\circ\)

Hence, part(ii) is proved, too.

Now, considering \(\triangle{DBC}\) and \(\triangle{ACB}\), we have,

AC = DB ...(from part(i))

Side BC is common.

and \(\angle{DBC}\) = \(\angle{ACB}\) = \(90^\circ\)

Therefore, \(\triangle{DBC}\) \(\displaystyle \cong \) \(\triangle{ACB}\) ...(SSA Congruency theorem)

Hence, now, part(iii) is proved, too.

Now, DC = AB ...(By CPCT)

Multipling both sides by 1/2, we get,

(1/2) DC = (1/2) AB

Now, as we know, CM = (1/2) DC

Therefore, CM = (1/2) AB

Hence, part (iv) is proved.

1. In an isosceles triangle ABC, with AB = AC, the bisectors of \(\angle{B}\) and \(\angle{C}\) intersect each other at O. Join A to O. Show that,

(i) OB = OC

(ii) AO bisects \(\angle{A}\)

(i) OB = OC

(ii) AO bisects \(\angle{A}\)

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In \(\triangle{ABC}\), we have

AB = AC ...(Given)

Also, \(\angle{B}\) = \(\angle{C}\) ...(Since corresponding angles of equal sides are equal)

Multipling both sides by 1/2, we get,

(1/2) \(\angle{B}\) = (1/2) \(\angle{C}\)

Therefore, \(\angle{OBC}\) = \(\angle{OCB}\)

Also, it is given that, OB and OC are bisectors of \(\angle{B}\) and \(\angle{C}\), respectively,

Therefore, \(\angle{OBA}\) and \(\angle{OCA}\)

Therefore, OB = OC ...(Since corresponding sides of equal angles are equal)

Hence, part (i) is proved.

In \(\triangle{ABO}\) and \(\triangle{ACD}\), we have,

AB = AC ...(Given)

\(\angle{OBA}\) = \(\angle{OCA}\) ...(from part(i))

OB = OC ...(proved earlier)

Therefore, \(\triangle{ABO}\) \(\displaystyle \cong \) \(\triangle{ACO}\) ...(By SAS congruency test)

Thus, \(\angle{BAO}\) = \(\angle{CAO}\) ...(By CPCT)

Hence, AO bisects \(\angle{A}\) is proved.

2. In the \(\triangle{ABC}\), AD is the perpendicular bisector of BC (see figure).

Show that \(\triangle{ABC}\) is an isosceles triangle in which AB = AC.

Show that \(\triangle{ABC}\) is an isosceles triangle in which AB = AC.

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In \(\triangle{ABC}\) and \(\triangle{ACD}\), we have,

DB = DC ...(given)

\(\angle{ADB}\) = \(\angle{ADC}\) ...(Since, AD is the perpendicular bisector of BC)

and AD is the Common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By SAS congruency test)

Therefore, AB = AC ...(By CPCT)

Hence, \(\triangle{ABC}\) is an isosceles triangle.

Hence, proved.

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides
AC and AB respectively (see figure).

Show that these altitudes are equal.

Show that these altitudes are equal.

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In \(\triangle{ABE}\) and \(\triangle{ACF}\) , we have,

\(\angle{AEB}\) = \(\angle{AFC}\) ...(BE and CF are perpendiculars drawn to sides AC and AB respectively)

Also, \(\angle{A}\) is common angle.

and AB = AC ...(Given)

Therefore, \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) ...(By AAS congruency test)

Therefore, AB = AC ...(By CPCT)

Hence, BE = CF ...(By CPCT)

Hence, proved.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see
figure). Show that

(i) \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\)

(ii) AB = AC i.e., ABC is an isosceles triangle.

(i) \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\)

(ii) AB = AC i.e., ABC is an isosceles triangle.

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In \(\triangle{ABE}\) and \(\triangle{ACF}\) , we have,

\(\angle{AEB}\) = \(\angle{AFC}\)(since, BE and CF are perpendicular to sides AC and AB)

\(\angle{BAE}\) = \(\angle{CAF}\) ...(since, \(\angle{A}\) is the Common angle)

and BE = CF ...(Given)

Therefore, \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) ...(By AAS Congruency test)

Thus, AB = AC ...(By CPCT)

So, \(\triangle{ABC}\) is an isosceles triangle.

Hence, proved.

5. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that \(\angle{ABD}\) = \(\angle{ACD}\).

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In \(\triangle{ABC}\), we have,

AB = AC ...(Since, \(\triangle{ABC}\) is an isosceles triangle)

Therefore, \(\angle{ABC}\) = \(\angle{ACB}\) ...(i)(Since, angles opposite to equal sides are equal)

Similarly, in \(\triangle{DBC}\), we have,

BD = CD ...(Since, \(\triangle{DBC}\) too, is an isosceles triangle)

Therefore, \(\angle{DBC}\) = \(\angle{DCB}\) ...(ii) (Since, angles opposite to equal sides are equal)

Now, On adding, Equations (i) and (ii), we get,

\(\angle{ABC}\) + \(\angle{DBC}\) = \(\angle{ACB}\) + \(\angle{DCB}\)

Therefore, \(\angle{ABD}\) = \(\angle{ACD}\)

Hence, proved.

6. \(\triangle{ABC}\) is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure).

Show that \(\angle{BCD}\) is a right angle.

Show that \(\angle{BCD}\) is a right angle.

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In \(\triangle{ABC}\), we have,

AB = AC ...(i)(Given)

\(\angle{ACB}\) = \(\angle{ABC}\) ...(ii)(Since, angles opposite to equal sides are equal)

Also, AB = AD ...(iii)(Given)

From (i) and (iii), AC = AD

Now, in \(\triangle{ADC}\) , we have,

AD = AC ...(proved earlier)

\(\angle{ACD}\) = \(\angle{ADC}\) ...(Since, angles opposite to equal sides are equal)

Also, \(\angle{ACD}\) = \(\angle{BDC}\) ...(iv)(Since, \(\angle{ADC}\) = \(\angle{BDC}\))

On adding Equations (ii) and (iv), we get,

\(\angle{ACB}\) + \(\angle{ACD}\) = \(\angle{ABC}\) + \(\angle{BDC}\)

i.e., \(\angle{BCD}\) = \(\angle{ABC}\) + \(\angle{BDC}\)

Adding \(\angle{BCD}\) on both sides, we have,

\(\angle{BCD}\) + \(\angle{BCD}\) = \(\angle{ABC}\) + \(\angle{BDC}\) + \(\angle{BCD}\)

i.e., 2 \(\angle{BCD}\) = \(180^\circ\) ...(Since, sum of all angles of a triangle is \(180^\circ\))

Therefore, \(\angle{BCD}\) = \(90^\circ\)

Hence, proved.

7. ABC is a right angled triangle in which \(\angle{A}\) = \(90^\circ\) and AB = AC, find \(\angle{B}\) and \(\angle{C}\).

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In \(\triangle{ABC}\), we have,

AB = AC ...(Given)

\(\angle{B}\) = \(\angle{C}\) ...(i)(Since, angles opposite to equal sides are equal)

Now, we know that,

\(\angle{A}\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\)

i.e., \(90^\circ\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\) ...(given)

i.e.,\(90^\circ\) + \(\angle{B}\) + \(\angle{B}\) = \(180^\circ\) ...(from(i))

i.e., 2 \(\angle{B}\) = \(180^\circ\) - \(90^\circ\)

i.e., 2 \(\angle{B}\) = \(90^\circ\)

Therefore, \(\angle{B}\) = \(45^\circ\)

Therefore, \(\angle{C}\) = \(45^\circ\), too.

8. Show that the angles of an equilateral triangle are \(60^\circ\) each.

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Let \(\triangle{ABC}\) be an equilateral triangle, such that

AB = BC = CA (by property)

Now, we have, AB = AC

\(\angle{B}\) = \(\angle{C}\) ...(i)(Since, angles opposite to equal sides are equal)

Similarly, as CB = CA

Thus, \(\angle{A}\) = \(\angle{B}\) ...(ii)(Since, angles opposite to equal sides are equal)

\(\angle{A}\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\) ...(iii)(Since, the sums of all angles of a triangle are \(180^\circ\))

From Equations (i),(ii) and (iii), we have,

\(\angle{A}\) + \(\angle{A}\) + \(\angle{A}\) = \(180^\circ\)

3 \(\angle{A}\) = \(180^\circ\)

\(\angle{A}\) = \(60^\circ\)

Therefore, \(\angle{A}\) = \(\angle{B}\) = \(\angle{C}\) = \(60^\circ\)

Hence, proved.

1.\(\triangle{ABC}\) and \(\triangle{DBC}\) are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that

(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\)

(ii)\(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\)

(iii) AP bisects \(\angle{A}\) as well as \(\angle{D}\)

(iv) AP is the perpendicular bisector of BC.

(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\)

(ii)\(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\)

(iii) AP bisects \(\angle{A}\) as well as \(\angle{D}\)

(iv) AP is the perpendicular bisector of BC.

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Given: \(\triangle{ABC}\) and \(\triangle{DBC}\) are two isosceles triangles having same base BC, such that AB = AC and BD = CD.

Proof: In \(\triangle{ABD}\) and \(\triangle{ACD}\), we have,

AB = AC ...(Given)

BD = CD ...(Given)

and AD is Common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By SSS Congruency test)

Hence, part (i) is proved.

In \(\triangle{ABP}\) and \(\triangle{ACP}\), we have,

AB = AC ...(Given)

Also, \(\angle{a}\) = \(\angle{b}\) ...(Since, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

and AP is the common side.

Therefore,
\(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\) ...(By SAS congruency test)

Hence, part (ii) is proved.

\(\angle{a}\) = \(\angle{b}\) ...(Since, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

Hence, proved that AP bisects \(\angle{A}\)

Now, \(\angle{ADB}\) = \(\angle{ACD}\) ...(Since, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\))

Now, by subtracting from \(180^\circ\), we get,

\(180^\circ\) - \(\angle{ADB}\) = \(180^\circ\) - \(\angle{ACD}\)

Therefore, \(\angle{BDP}\) = \(\angle{ACD}\) ...(By linear pair axiom)

Hence, proved that AP bisects \(\angle{D}\)

Hence, part (iii) is proved.

Now, BP = CP (Since, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\))

and \(\angle{c}\) = \(\angle{d}\) ...(i)

But \(\angle{c}\) + \(\angle{d}\) = \(180^\circ\) ...(Linear pair axiom)

i.e., \(\angle{c}\) + \(\angle{c}\) = \(180^\circ\) ...(from (i))

i.e., 2 \(\angle{c}\) = \(180^\circ\)

Therefore, \(\angle{c}\) = \(90^\circ\)

Therefore, \(\angle{c}\) = \(\angle{d}\) = \(90^\circ\)

Therefore, AP is the perpendicular bisector of BC.

Hence, proved.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects \(\angle{A}\)

(i) AD bisects BC

(ii) AD bisects \(\angle{A}\)

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In \(\triangle{ABD}\) and \(\triangle{ACD}\), we have,

AB = AC ...(Given)

\(\angle{ABD}\) = \(\angle{ADC}\) = \(90^\circ\) ...(Since AD is an altitude)

and AD is Common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By RHS congruency axiom)

Therefore, BD = DC ...(By CPCT)

Thus, we can say that, AD bisects BC.

Also, \(\angle{BAD}\) = \(\angle{CAD}\) ...(By CPCT)

Thus, this shows that AD bisects \(\angle{A}\).

Hence, proved.

3. The two sides AB and BC and median AM of one triangle ABC are respectively
equal to sides PQ and QR and median PN of \(\triangle{PQR}\) (see figure). Show that

(i) \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\)

(ii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\)

(i) \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\)

(ii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\)

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We have, AM is the median of \(\triangle{ABC}\),

So, BM = MC = (1/2) BC ...(i)(Since, M Bisects BC)

Similarly, PN is the median of \(\triangle{PQR}\).

So, QN = NR = (1/2) QR ...(ii)(Since, N Bisects QR)

Now, we have,

BC = QR ...(Given)

Multiplying both sides by 1/2, we get,

(1/2) BC = (1/2) QR

Therefore, BM = QN ...(iii)(From eq.(i) and (ii))

In \(\triangle{ABM}\) and \(\triangle{PQN}\), we have,

AB = PQ ...(Given)

AM = PN ...(Given)

and BM = QN ...(From Eq.(iii))

\(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\) ...(By SSS congruency test)

Hence, part (i) is proved.

In \(\triangle{ABC}\) and \(\triangle{PQR}\), we have,

AB = PQ ...(Given)

\(\angle{B}\) = \(\angle{Q}\) ...(Since, \(\triangle{ABM}\) \(\displaystyle \cong \) \(\triangle{PQN}\))

and BC = QR ...(Given)

\(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\) ...(By SAS congruency test)

Hence, proved.

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

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In \(\triangle{BEC}\) and \(\triangle{CBF}\), we have,

\(\angle{BEC}\) = \(\angle{CFB}\) = \(90^\circ\) ...(Given)

BE = CF ...(Given)

and
BC is the Common side.

Therefore, \(\triangle{BEC}\) \(\displaystyle \cong \) \(\triangle{CFB}\) ...(By RHS congruency test)

Thus, EC = FB ...(i)(By CPCT)

In \(\triangle{AEB}\) and \(\triangle{AFC}\), we have,

\(\angle{AEB}\) = \(\angle{AFC}\) = \(90^\circ\) ...(Given)

\(\angle{A}\) is Common angle.

and EB = FC ...(Given)

Therefore, \(\triangle{AEB}\) \(\displaystyle \cong \) \(\triangle{AFC}\) ...(By AAS congruency test)

Thus, AE = AF ...(ii)(By CPCT)

Now, on adding Equation (i) and (ii), we get,

EC + AE = FB + AF

AC = AB ...(Since, AC = EC + AE and AB = FB + AF)

Hence, it is proved that, \(\triangle{ABC}\) is an isosceles triangle.

5. ABC is an isosceles triangle with AB = AC. Draw AP perpendicular to BC to show that \(\angle{B}\) = \(\angle{C}\)

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In \(\triangle{ABP}\) and \(\triangle{ACP}\), we have,

AB = AC ...(Given)

AP is the Common side.

\(\angle{ABP}\) = \(\angle{ACP}\) = \(90^\circ\) ...(Since, AP perpendicular to BC)

Therefore, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\) ...(By RHS congruency test)

Thus, \(\angle{B}\) = \(\angle{C}\) ...(By CPCT)

Hence, proved.

1. Show that in a right angled triangle, the hypotenuse is the longest side.

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Let ABC be a right angled triangle, such that \(\angle{ABC}\) = \(90^\circ\).

We know that, \(\angle{ABC}\) + \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\) ...(Since, the sums of all angles of a triangle are \(180^\circ\))

\(90^\circ\) + \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\)

\(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\) - \(90^\circ\)

\(\angle{BCA}\) + \(\angle{CAB}\) = \(90^\circ\)

Thus, we can say that, both the angles \(\angle{BCA}\) and \(\angle{CAB}\) are acute.

\(\angle{BCA}\) < \(90^\circ\)

and \(\angle{CAB}\) < \(90^\circ\)

Thus, \(\angle{BCA}\) < \(\angle{ABC}\) and \(\angle{CAB}\) < \(\angle{ABC}\)

Therefore, we can say,

AB < AC and BC < AC ...(Since, side opposite to greater angle is longer)

Hence, the hypotenuse AC is the longest side is proved.

2. In figure, sides AB and AC of are extended to points P and Q respectively. Also, \(\angle{PBC}\) < \(\angle{QCB}\). Show that AC > AB.

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By linear pair axiom, we say that,

\(\angle{ACB}\) + \(\angle{QCB}\) = \(180^\circ\) ...(i)

\(\angle{ABC}\) + \(\angle{PBC}\) = \(180^\circ\) ...(ii)

From Equation (i) and (ii), we get,

\(\angle{ACB}\) + \(\angle{QCB}\) = \(\angle{ABC}\) + \(\angle{PBC}\)

But \(\angle{PBC}\) < \(\angle{QCB}\) ...(given)

Therefore, \(\angle{ABC}\) > \(\angle{ACB}\)

Therefore, AC > AB.

Hence, proved.

3. In figure, \(\angle{B}\) < \(\angle{A}\) and \(\angle{C}\) < \(\angle{D}\). Show that AD < BC.

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We have, \(\angle{B}\) < \(\angle{A}\) ...(given)

Also, \(\angle{C}\) < \(\angle{D}\) ...(given)

Since, side opposite to greater angle is longer

we get, AO < BO ...(i)

similarly, OD < OC ...(ii)

On adding, Equation (i) and (ii), we have,

AO + OD < BO + OC

i.e., AD < BC

Hence, proved.

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD
(see figure). Show that \(\angle{A}\) > \(\angle{C}\) and \(\angle{B}\) > \(\angle{D}\).

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Given: ABCD is a quadrilateral having AB as the smallest side and CD as the longest side

To prove: \(\angle{A}\) > \(\angle{C}\) and \(\angle{B}\) > \(\angle{D}\)

Construction: Join A to C and B to D.

Figure:

Proof: In \(\triangle{ABC}\), we have AB is the smallest side.

As, AB < BC

Thus, \(\angle{5}\) < \(\angle{1}\) ...(i)(Since, angle opposite to longer side is greater)

Similarly, in \(\triangle{ADC}\), we have CD is the largest side.

As, AD < CD

Thus, \(\angle{6}\) < \(\angle{2}\) ...(ii)(Since, angle opposite to longer side is greater)

On adding Equation (i) and (ii), we get,

i.e., \(\angle{5}\) + \(\angle{6}\) < \(\angle{1}\) + \(\angle{2}\)

Therefore, \(\angle{C}\) < \(\angle{A}\)

i.e., \(\angle{A}\) > \(\angle{C}\)

Now, Similarly, \(\triangle{ADB}\), we have AB is the smallest side.

As, AD > AB

Thus, \(\angle{3}\) > \(\angle{8}\) ...(iii)(Since, angle opposite to longer side is greater)

Similarly, in \(\triangle{BCD}\), we have CD is the largest side.

As, CD > BC

Thus, \(\angle{4}\) > \(\angle{7}\) ...(iv)(Since, angle opposite to longer side is greater)

On adding Equation (iii) and (iv), we get,

i.e., \(\angle{3}\) + \(\angle{4}\) > \(\angle{8}\) + \(\angle{7}\)

Therefore, \(\angle{B}\) > \(\angle{D}\)

Hence, proved.

5. In figure, PR > PQ and PS bisects \(\angle{QPR}\). Prove that \(\angle{PSR}\) > \(\angle{PSQ}\).

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In \(\triangle{PQR}\), we have,

PR > PQ ...(given)

Therefore, \(\angle{PQR}\) > \(\angle{PRQ}\) ...(i)(Since, angle opposite to longer side is greater)

We can also say,

\(\angle{a}\) = \(\angle{b}\) ...(ii)(since, PS bisects \(\angle{QPR}\))

On adding Equation (i) and (ii), we get,

\(\angle{PQR}\) + \(\angle{a}\)> \(\angle{PRQ}\) + \(\angle{b}\) ...(iii)

Also, \(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(180^\circ\) ...(iv)

and \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\) = \(180^\circ\) ...(v)(Since, the sums of all angles of a triangle are \(180^\circ\))

From Equation (iv) and (v), we get,

\(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\)

Since, \(\angle{PQS}\) = \(\angle{PQR}\) and \(\angle{PRS}\) = \(\angle{PRQ}\), we get,

\(\angle{PQR}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRQ}\) + \(\angle{PSR}\) + \(\angle{b}\)

Rearranging the terms, we get,

\(\angle{PQR}\) + \(\angle{a}\) + \(\angle{PSQ}\) = \(\angle{PRQ}\) + \(\angle{b}\) + \(\angle{PSR}\) ...(vi)

From Equation (iii) and (vi), we get,

\(\angle{PSQ}\) < \(\angle{PSR}\) ...(Since, side opposite to greater angle is longer)

Hence, proved.

6. Show that of all line segments drawn from a give point not on it, the perpendicular
line segment is the shortest.

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Given: x is a line and A is a point not lying on x.

C is any point on x other than B.

To prove: AB < AC

Proof: In \(\triangle{ABC}\), \(\angle{B}\) is the right angle.

Therefore, \(\angle{C}\) is an acute angle.

i.e., , \(\angle{C}\) < \(\angle{B}\)

Thus, AB < AC ....(Since, side opposite to greater angle is longer)

Hence, proved that the perpendicular line segment is the shortest.