In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.
In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.
Solution and Explanation
It is given that:
\[ \angle EPA = \angle DPB \]
By angle addition property:
\[ \angle EPA + \angle DPE = \angle DPB + \angle DPE \]
Therefore, we have:
\[ \angle DPA = \angle EPB \]
In \( \triangle DAP \) and \( \triangle EBP \):
\[ \angle DAP = \angle EBP \quad \text{(Given)} \]
\[ AP = BP \quad \text{(P is the mid-point of AB)} \]
\[ \angle DPA = \angle EPB \quad \text{(From above)} \]
By the ASA congruence rule:
\[ \triangle DAP \cong \triangle EBP \]
By CPCT (Corresponding Parts of Congruent Triangles):
\[ AD = BE \]
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