Question

if 2 isosceles triangles have a common base, prove that line joining the vertices bisects the base at right angle .

Answer 1

Answer:

Step-by-step explanation:

ABC and DBC are two isosceles triangles with the common base, BC.

In ABD and ACD

AB = AC;

BD = CD;

AD = AD

ABD  ACD             (SSS congruence rule)

 

BAE = CAE                    (c.p.c.t)

In ABE and ACE

AB = AC;

BAE = CAE ;

   

 

AE = AE

 

ABE  ACE                                        (SAS congruence rule)

BEA = CEA and BE = EC                   (c.p.c.t)

BEA + CEA = 180o                                 (linear pair)

2BEA = 180o

 BEA = 90o

BEA = 90o = CEA

 

Hence, AD bisects the base BC at right angles.

Answer 2

Answer:

Step-by-step explanation:

ABC and DBC are two isosceles triangles with the common base, BC.

In ABD and ACD

AB = AC;

BD = CD;

AD = AD

ABD ACD (SSS congruence rule)

BAE = CAE (c.p.c.t)

In ABE and ACE

AB = AC;

BAE = CAE ;

AE = AE

ABE ACE (SAS congruence rule)

BEA = CEA and BE = EC (c.p.c.t)

BEA + CEA = 180o (linear pair)

2BEA = 180o

BEA = 90o

BEA = 90o = CEA

Hence, AD bisects the base BC at right angles.