Question

AX is the bisector of ZBAC, P is any point on AX. Show that the perpendiculars. drawn from P to AB and AC are equal​

Answer 1

$\tt \:In \: ∆APM \: and \: ∆APN \\ \tt \: < 1 = < 2 \: (given)\\ \tt \:AP=AP(Common)\\ \tt \: <M=<N=90°\\ \tt \: By \: ASA \: congruence \: rule \\ \tt \: ∆APM \: ≈ \: ∆APN\\ \tt \:MP=PN(CPCT)$

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$\sf \: @WildCat7083$

Answer 2

Given:

AX is the bisector of ∠BAC

P- point on AX

To prove:

MP=PN

Proof:

Consider ΔAMP and ΔANP

AX is the bisector of ∠BAC. So, ∠MAP = ∠PAN

AP=AP (Common)

∠AMP = ∠ANP = 90°

SO, from ASA congruency, ΔAMP ≅ ΔANP

∴MP=PN (Corresponding Parts of Congruent Triangles are equal)

Hence, proved.