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
You want to show that ΔART ≅ ΔPEN,
If it is given that AT = PN and you are to use ASA criterion, you need to have
1) ?
2) ?
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.
In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE. 
In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.
Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).
PQRS is a parallelogram. L and M are points on PQ and SR respectively such that PL = MR.
Show that LM and QS bisect each other.
ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
