This is very important question of ncert class 9^{th} of chapter triangles. How I solve the best solution of exercise 7.1 question number 6. Please help me to solve this in a easy and best way ad give me the simple method of this question. In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.

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# In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. Q.6

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Solution:It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC

To prove:The line segment BC and DE are similar i.e. BC = DE

Proof:We know that BAD = EAC

Now, by adding DAC on both sides we get,

BAD + DAC = EAC +DAC

This implies, BAC = EAD

Now, ΔABC and ΔADE are similar by SAS congruency since:

(i) AC = AE (As given in the question)

(ii) BAC = EAD

(iii) AB = AD (It is also given in the question)

∴ Triangles ABC and ADE are similar i.e. ΔABC ΔADE.

So, by the rule of CPCT, it can be said that BC = DE.