maths

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.**

Let D be the mid-point on BC. $DF⊥AC$ and $DE⊥AB$

Now, In $△BED$ and $△CFD$

$∠DEB=∠DFC$ (Each $90_{∘}$)

$BD=CD$ (D is mid point of BC)

$∠B=∠C$ (Given)

thus, $△DBE≅△DCF$ (ASA rule)

Hence, $DE=DF$ (By cpct)

Hence proved.

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