Question

In any triangle ABC, BP and CQ are perpendicular to any line
through A and M is the mid-point of BC.
Show that MP = MQ​

Answer 1

Answer:

Step-by-step explanation:

Let the perpendicular from D to AB meet at O and AC meet at P

Then in ∆ AOD and ∆ APD

angle P= angle O= 90°

Given DO=DP

AD common

So ∆AOD=∆APD (As per RHS )

As per CPCT

AO=AP eq-1

Same we can prove ∆DOB=∆DOC

And OB=OC eq-2

Adding the eq-1 and eq-2 we get

AO+OB=AO+OC

=>AB=AC

Answer 2

Answer:v

In the given figure,  

M is the midpoint of BC, that means = BM = MC.

Also, BP ⊥⊥ AQ , CQ ⊥⊥ AQ

​From point M, draw MX ⊥⊥ AQ

​Since BP, MX and CQ are perpendicular to same line AQ, so they are parallel to each other.

Since M is the midpoint of BC, so intercepts made by these parallel lines on PQ will also be equal.

Hence, PX = XQ.

In triangle MPX and MQX,

PX = XQ    [proved above]

∠∠PXM = ∠∠QXM   [each 900 ]

MX = MX   [common]

so, triangle MPX is congruent to triangle MQX [SAS]

so, MP = MQ = [CPCT]

Step-by-step explanation: