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State if the following is true or false.

In the given figure $DB ⊥ BC, DE ⊥ AB$ and $AC ⊥ BC,$ then $Δ BDE ~ Δ ABC$ .

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True

False

answer is A.

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Detailed Solution

It is given that the sides

D B ⊥ B C

, DE 丄AB and the sides

A C ⊥ B C

In the given figure DB BC, DE AB and AC BC . Prove that BEDE = ACBC

So, the angle B and angle C will be

90 °

.
Then,

∠ B + ∠ C = 90 ° + 90 ° ⇒ ∠ B + ∠ C = 180 °

Therefore ,

B D ∥ A C

.
Since the angles

∠ E B D

and

∠ C A B

are alternate angles and the line AB crosses both the lines BD and AC and

B D ∥ A C

, this implies that

∠ E B D = ∠ C A B

.
Now consider the triangles BDE and ABC.
It is clear from the figure that the angles E and angle C are right angles,
∠DEB=∠BCA=90°
Also, the angles EBD and CAB are the alternative angles that are

∠ E B D = ∠ C A B

. Hence,

Δ B D E

and

Δ A B C

are similar.
Therefore, the statement is true.
Option 1 is correct.

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