in the figure abc is a right triangle right angled at b the bisectors of angle a and angle c meet at angle d find angle adc
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Step-by-step explanation:
In ΔABD and ΔACD
AB=AC (given)
∠BAD=∠CAD
As AD is bisector of ∠A and AD=AD
ΔDAB=ΔDAC (by SAS congruency rule)
∠ADB=∠ADC (by c.p.c.t)
∠ADB=∠ADC=90
0
and BD=DC
In ΔABD,
AD
2
+BD
2
=AB
2
…(i)
AD
2
+DC
2
=AC
2
…(ii)
Adding (i) and (ii), we get
2AD
2
+BD
2
+DC
2
=AB
2
+AC
2
2AD
2
+BD
2
+DC
2
=BC
2
2AD
2
+2BD
2
=BC
2
2(AD
2
+BD
2
)=BC
2
2[AD
2
+(
2
1
BC)
2
]=BC
2
2[AD
2
+
4
1
BC
2
]=BC
2
2AD
2
+
2
1
BC
2
=BC
2
2AD
2
=BC
2
−
2
1
BC
2
=
2
BC
2
4AD
2
=BC
2
2AD=BC
i.e. BC=2AD
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