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Answer :

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.

- \(\triangle{ABC}\) 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.
- AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects \(\angle{A}\)
- 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) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{PQR}\)
- 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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