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AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠ A.
(i) In ΔBAD and ΔCAD,
∠ADB = ∠ADC (Each 90º as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
∴ΔBAD ≅ ΔCAD (By RHS Congruence rule)
⇒ BD = CD (By CPCT)
Hence, AD bisects BC.
(ii) Also, by CPCT,
∠BAD = ∠CAD
Hence, AD bisects ∠A.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
In ΔBEC and ΔCFB,
∠BEC = ∠CFB (Each 90°)
BC = CB (Common)
BE = CF (Given)
∴ ΔBEC ≅ ΔCFB (By RHS congruency)
⇒ ∠BCE = ∠CBF (By CPCT)
∴ AB = AC (Sides opposite to equal angles of a triangle are equal)
Hence, ΔABC is isosceles.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.
In ΔAPB and ΔAPC,
∠APB = ∠APC (Each 90º)
AB =AC (Given)
AP = AP (Common)
∴ ΔAPB ≅ ΔAPC (Using RHS congruence rule)
⇒ ∠B = ∠C (By using CPCT)
Triangle ABC and triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the
same side of BC (see Figure). If AD is extended to intersect BC at P, show that
(i) triangleABD ≅
triangle ACD
(ii) triangle ABP ≅ triangle ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
(i) In ΔABD and ΔACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common)
∴ ΔABD ≅ ΔACD (By SSS congruence rule)
⇒ ∠BAD = ∠CAD (By CPCT)
⇒ ∠BAP = ∠CAP …. (1)
(ii) In ΔABP and ΔACP,
AB = AC (Given)
∠BAP = ∠CAP [From equation (1)]
AP = AP (Common)
∴ ΔABP ≅ ΔACP (By SAS congruence rule)
⇒ BP = CP (By CPCT) … (2)
(iii) From equation (1),
∠BAP = ∠CAP
Hence, AP bisects ∠A.
In ΔBDP and ΔCDP,
BD = CD (Given)
DP = DP (Common)
BP = CP [From equation (2)]
∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule)
⇒ ∠BDP = ∠CDP (By CPCT) … (3)
Hence, AP bisects ∠D.
(iv) ΔBDP ≅ ΔCDP
∴ ∠BPD = ∠CPD (By CPCT) …. (4)
∠BPD + ∠CPD = 180
(Linear pair angles)
∠BPD + ∠BPD = 180
2∠BPD = 180 [From equation (4)]
∠BPD = 90 … (5)
From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of â?? PQR (see Figure). Show that:
(i) triangle ABM ≅ triangle PQN
(ii) triangle ABC ≅ triangle PQR
(i) In ΔABC, AM is the median to BC.
∴ BM = 
BC
In ΔPQR, PN is the median to QR.
∴ QN = QR
However, BC = QR
∴ BC = QR
⇒ BM = QN … (1)
In ΔABM and ΔPQN,
AB = PQ (Given)
BM = QN [From equation (1)]
AM = PN (Given)
∴ ΔABM ≅ ΔPQN (SSS congruence rule)
∠ABM = ∠PQN (By CPCT)
∠ABC = ∠PQR … (2)
(ii) In ΔABC and ΔPQR,
AB = PQ (Given)
∠ABC = ∠PQR [From equation (2)]
BC = QR (Given)
⇒ ΔABC ≅ ΔPQR (By SAS congruence rule)
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