Question

The perpendiculars from the mid-point of BC to AB and AC are equal.

If the above statement is true then mention answer as 1, else mention 0 if false.

Answer 1

Given: $△ABC$, $\angle B=\angle C$
Let D be the mid-point on BC. $DF\perp AC$ and $DE\perp AB$
Now, In $△BED$ and $△CFD$
$\angle DEB=\angle DFC$ (Each ${90}^{\circ }$)
$BD=CD$ (D is mid point of BC)
$\angle B=\angle C$ (Given)
thus, $△DBE\cong △DCF$ (ASA rule)
Hence, $DE=DF$ (By cpct)
Hence proved.