Advertisements
Advertisements
Question
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 ).
Advertisements
Solution
ΔABC is an equilateral triangle,
So, each of its angles equals 60°.
QP is parallel to AC,
⇒ ∠PQB = ∠RAQ = 60°
ln ΔQBP,
∠PQB = ∠BQP = 60°
So, ∠PBQ + ∠BQP + ∠BPQ = 180° ....(angle sum property)
⇒ 60°+ 60° + ∠BPQ = 180°
⇒ ∠BPQ = 60°
So, ΔBPQ is an equilateral triangle.
⇒ QP = BP
⇒ QP = CR ....(i)
Now, ∠QPM + ∠BPQ = 180° ...(linear pair)
⇒ ∠QPM+ 60°= 180°
⇒ ∠QPM = 120°
Also, ∠RCM+ ∠ACB = 180° ...(linear pair)
⇒ ∠RCM+ 60° = 180°
⇒ ∠RCM = 120°
ln ΔRCM and ΔQMP,
∠RCM = ∠QPM ....(each is 120°)
∠RMC = ∠QMP ...(vertically opposite angles)
QP= CR ....(from(i))
⇒ ΔRCM ≅ ΔQMP ....(AAS congruence criterion)
So, CM = PM
⇒ QR bisects PC.
APPEARS IN
RELATED QUESTIONS
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
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.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from B and C to the opposite sides are equal.
In ∆ABC, AB = AC. Show that the altitude AD is median also.
In the following figure, BL = CM.
Prove that AD is a median of triangle ABC.
In the following diagram, ABCD is a square and APB is an equilateral triangle.
- Prove that: ΔAPD ≅ ΔBPC
- Find the angles of ΔDPC.
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.
