Question

5. A triangle ABC has angleB = angleC.
Prove that:
(i) the perpendiculars from the mid-point of
BC to AB and AC are equal.
(ii) the perpendiculars from B and C to the
opposite sides are equal.​

Answer 1

Step-by-step explanation:

Given:−

In △ABC, ∠B=∠C                  

DL is perpendicular from D to AB

DM is the perpendicular from D to AC

Toprove:−

DL=DM

Proof:−

In △DLB and △DMC

⇒  ∠DLB=∠DMC=90  

o

        [ DL⊥AB and DM⊥AC ]

⇒  ∠B=∠C                    [ Given ]

⇒  BD=AC                    [ D is the mid point of BC ]

∴  △DLB≅△DMC           [ By AAS criteria of congruence ]

∴  DL=DM                   [ c.p.c.t ]

(ii)

Given:−

In △ABC, ∠B=∠C.      

BP is perpendicular from D to AC

CQ is the perpendicular from C to AB

Toprove:−

BP=CQ

Proof:−

In △BPC and △CQB

⇒  ∠B=∠C                        [ Given ]

⇒  ∠BPC=∠CQB=90  

o

               [ BP⊥AC and CQ⊥AB ]

⇒  BC=BC                           [ Common side ]

⇒  △BPC≅△CQB        [ By AAS criteria ]

∴  BP=CQ          [ c.p.c.t ]  

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