Question

A triangle ABC has angleB=angleC. prove that
1. the perpandicular from the midpoint of BC on AB and AC are equal.
2. the perpandicular form B and C to the opposite sides are equal.​

Answer 1

Answer:

Step-by-step explanation:(i)  

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 ]    

Answer 2

Step-by-step explanation:

(i)

Given:−

In △ABC, ∠B=∠C

DL is perpendicular from D to AB

DM is the perpendicular from D to AC

To prove:−

DL=DM

Proof:−

In △DLB and △DMC

⇒ ∠DLB=∠DMC=90

[ DL⊥AB and DM⊥AC ]

⇒ ∠B=∠C

⇒ BD=AC

∴ △DLB≅△DMC

∴ DL=DM

(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

⇒ ∠BPC=∠CQB=90

[ BP⊥AC and CQ⊥AB ]

⇒ BC=BC

⇒ △BPC≅△CQB

∴ BP=CQ

$\huge{ son \: prodigal}$