In $Δ A B C$ , AB = BC = 6 cm. If a circle is drawn with center at B and radius 2 cm which intersects AB and BC at E and D, then show that $Δ A B C \sim Δ E B D$ .

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Solution

In $Δ A B C a n d Δ E B D$ ,

$\frac{B E}{B A} = \frac{B D}{B C} = \frac{2}{6} = \frac{1}{3} a n d ∠ A B C = ∠ E B D$

$∴$ By SAS similarity, $Δ A B C$ and $Δ E B D$ are similar.

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In ΔABC, AB = BC = 6 cm. If a circle is drawn with center at B and radius 2 cm which intersects AB and BC at E and D, then show that ΔABC∼ΔEBD.

In ΔABC and ΔEBD, BEBA=BDBC=26=13 and ∠ABC=∠EBD ∴ By SAS similarity, ΔABC and ΔEBD are similar.

In $Δ A B C$ , AB = BC = 6 cm. If a circle is drawn with center at B and radius 2 cm which intersects AB and BC at E and D, then show that $Δ A B C \sim Δ E B D$ .

In $Δ A B C a n d Δ E B D$ ,

$\frac{B E}{B A} = \frac{B D}{B C} = \frac{2}{6} = \frac{1}{3} a n d ∠ A B C = ∠ E B D$

$∴$ By SAS similarity, $Δ A B C$ and $Δ E B D$ are similar.