in the figure,AB=AC,D is a pint on AC and E is a point on AB such that AD=ED=EC=BC. prove that angle A: angle B=1:3
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Rearrenging the given relation
BC×BC=AC×AD We can write
BCCD=ACBC→ΔABC similar ΔBDC
Their corresponding angle pairs are:
1.∠BAC= corresponding ∠DBC
2.∠ABC= corresponding ∠BDC
3.∠ACB =corresponding ∠DCB
So as per above relation 2 we have
∠ABC= corresponding ∠BDC
Again inΔABC
AB=AC→∠ABC=∠ACB=∠DCB
∴In ΔBDC,∠BDC=∠BCD
→BD=BC
Alternative way
The ratio of corresponding sides may be written in extended way as follows
BCCD=ACBC=ABBD
From this relation we have
ACBC=ABBD
⇒ACBC=ACBD→As AB=AC given
⇒1BC=1BD
⇒BC=BD
Proved
BC×BC=AC×AD We can write
BCCD=ACBC→ΔABC similar ΔBDC
Their corresponding angle pairs are:
1.∠BAC= corresponding ∠DBC
2.∠ABC= corresponding ∠BDC
3.∠ACB =corresponding ∠DCB
So as per above relation 2 we have
∠ABC= corresponding ∠BDC
Again inΔABC
AB=AC→∠ABC=∠ACB=∠DCB
∴In ΔBDC,∠BDC=∠BCD
→BD=BC
Alternative way
The ratio of corresponding sides may be written in extended way as follows
BCCD=ACBC=ABBD
From this relation we have
ACBC=ABBD
⇒ACBC=ACBD→As AB=AC given
⇒1BC=1BD
⇒BC=BD
Proved
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