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Answer :

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,

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

\(\Rightarrow \) \(\angle{BAC}\) = \(\angle{DAE}\)

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

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

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

Hence, proved.

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

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