Advertisements
Advertisements
प्रश्न
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
Advertisements
उत्तर
It is given that
∠B = 35°
∠C = 65°
AP is the bisector of ∠CAB

We have to arrangeAP, BPand CPin descending order.
In ΔACP we have
∠ACP = 65°
∠CAP = 40°(As AP is the bisector of ∠CAB
So AP > CP (Sides in front or greater angle will be greater) ........(1)
In ΔABP we have
∠BAP = 40°(As AP is the bisector of ∠CAB)
Since,
∠BAP >∠ABP
So BP > AP ..........(2)
Hence
From (1) & (2) we have
BP > AP > CP
APPEARS IN
संबंधित प्रश्न
You have to show that ΔAMP ≅ AMQ.
In the following proof, supply the missing reasons.
| Steps | Reasons | ||
| 1 | PM = QM | 1 | ... |
| 2 | ∠PMA = ∠QMA | 2 | ... |
| 3 | AM = AM | 3 | ... |
| 4 | ΔAMP ≅ ΔAMQ | 4 | ... |

In Fig. 10.40, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ ΔSAT.
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
In Fig. 10.99, AD ⊥ CD and CB ⊥. CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
(ii) AO bisects angle BAC.
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 a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
