Advertisements
Advertisements
प्रश्न
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.
Advertisements
उत्तर
We have, l is the bisector of ∠QAP.
∴ ∠QAB = ∠PAB
and ∠Q = ∠P ...[Each 90°]
⇒ ∠ABQ = ∠ABP ...[By angle sum property of △]
i. Now, in △APB and △AQB, we have
∠ABP = ∠ABQ ...[Proved above]
AB = BA ...[Common]
∠PAB ≅ ∠QAB ...[Given]
△APB ≅ △AQB ...[By ASA congruency]
ii. Since, △APB ≅ △AQB
⇒ BP = BQ ...[By Corresponding parts of congruent triangles]
i.e., [Perpendicular distance of B from AP] = [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A.
APPEARS IN
संबंधित प्रश्न
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
Explain, why ΔABC ≅ ΔFED.
In Fig. 10.92, it is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
Which of the following statements are true (T) and which are false (F):
If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
In the given figure, prove that:
CD + DA + AB + BC > 2AC
Use the information in the given figure to prove:
- AB = FE
- BD = CF
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB is parallel to EC.
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that: BD = CD
Which of the following is not a criterion for congruence of triangles?
