Question

In the adjoining figure BP I AC, CQ I AB
A-P-C, A-Q-B , then prove that
A APB and A AQC are similar.
Solution :
In A APB and A AQC
Z APB = 90° (1)
ZAQC = 90° (m​

Answer 1

Given :

In the adjoining figure, BP ⊥ AC, CQ ⊥ AB, A – P – C, A – Q – B, then prove that ∆APB and ∆AQC are similar.  

To Find :  Show that ΔAPB and ΔAQC are similar.

Solution:

ΔAPB and ΔAQC

∠A = ∠ A   Common

∠APB = ∠AQC = 90°

=> ΔAPB ≈ ΔAQC  (AA)

QED

Hence proved

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Answer 2

For a given correspondence, if two pairs of corresponding sides are in the same proportion and the angle between them is congruent, then the two triangles are similar. In the given figure, if AB/PQ = BC/QR, and ∠B ≅∠Q, then ∆ABC ~ ∆PQR 3. SSS test for similarity of triangles: For a given correspondence, if three sides of one triangle are in proportion with the corresponding three sides of the another triangle, then the two triangles are similar. In the given figure, of AB/PQ = BC = QR = AC/PR, then ∆ABC ~ ∆PQR Properties of similar triangles: 1. Reflexivity: ∆ABC ~ ∆ABC 2. Symmetry : If ∆ABC ~ ∆DEF, then ∆DEF ~ ∆ABC. 3. Transitivity: If ∆ABC ~ ∆DEF and ∆DEF ~ ∆GHI, then ∆ABC ~ ∆GHI.Read more on Sarthaks.com - https://www.sarthaks.com/849902/in-the-adjoining-figure-bp-ac-cq-ab-a-p-c-a-q-b-then-prove-that-apb-and-aqc-are-similar