Advertisements
Advertisements
प्रश्न
Q is a point on the side SR of a ∆PSR such that PQ = PR. Prove that PS > PQ.
Advertisements
उत्तर
In triangle PSR, Q is a point on the side SR such that PQ = PR.
To proof that PS > PQ

Proof: In triangle PRQ,
PQ = PR ...[Given]
∠R = ∠PQR ...(i) [Angle opposite to equal sides are equal]
∠PQR > ∠S ...(ii) [Exterior angle of a triangle is greater than each of the opposite interior angle]
Now, from equation (i) and (ii), we get
∠R > ∠S
PS > PR ...[Side opposite to greater angle is longer]
PS > PQ ...[PQ = PR]
APPEARS IN
संबंधित प्रश्न
Is the following statement true and false :
An exterior angle of a triangle is greater than the opposite interior angles.
In the given figure, AE bisects ∠CAD and ∠B= ∠C. Prove that AE || BC.

If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4 : 5. The angles of the triangle are
Can a triangle together have the following angles?
85°, 95° and 22°
Classify the following triangle according to sides:

In the given figure, AB is parallel to CD. Then the value of b is
S is any point on side QR of a ∆PQR. Show that: PQ + QR + RP > 2PS.
In the following figure, AB = BC and AD = BD = DC. The number of isosceles triangles in the figure is ______.

Can we have two acute angles whose sum is a straight angle? Why or why not?
