In a triangle PQR, N is a point on PR such that QN ⊥ PR . If PN. NR = QN², prove that ∠PQR = 90°
Solution:
Given, N is a point on PR of a triangle PQR such that QN⊥PR
Also, PN . NR = QN²
We have to prove that ∠PQR = 90°
PN . NR = QN² can be written as
PN/QN = QN/NR
From △PNQ and △RNQ,
PN/QN = QN/NR
∠PNQ = ∠RNQ = 90°
SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.
By SAS criterion, the triangles PNQ and RNQ are similar.
By the property of similar triangles,
∠PQN = ∠ QRN -------------- (1)
∠RQN = ∠QPN --------------- (2)
Adding (1) and (2),
∠PQN + ∠RQN = ∠QRN + ∠QPN
From the figure,
∠PQR = ∠PQN + ∠RQN ----------- (3)
So, ∠PQR = ∠QRN + ∠QPN --- (4)
We know that the sum of all the three interior angles of a triangle is always equal to 180°
In △QNR,
∠QNR + ∠NRQ + ∠RQN = 180°
90°+ ∠NRQ + ∠RQN= 180°
∠NRQ + ∠RQN = 180° - 90°
∠NRQ + ∠QPN = 90° --- [From (2)]
∠PQR = 90° [From (4)]
Therefore, it is proved that ∠PQR = 90°
✦ Try This: A and B are respectively the points on the sides PQ and PR of a ΔPQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR ? Give reason for your answer
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.3 Problem 11
In a triangle PQR, N is a point on PR such that QN ⊥ PR . If PN. NR = QN², prove that ∠PQR = 90°
Summary:
In a triangle PQR, N is a point on PR such that QN ⊥ PR . If PN. NR = QN², it is proved that ∠PQR = 90°
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