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.
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Given: △ABC, ∠B=∠C 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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