Two sides AB and AC and median AM of one triangle ABC are respectively equal to the sides PQ and QR and median PN of triangle PQR.
show that : (I) triangle ABM is congruent to triangle PQN
(II)triangle ABC is congruent to triangle PQR
Two sides AB and AC and median AM of one triangle ABC are respectively equal to the sides PQ and QR and median PN of triangle PQR.
show that : (I) triangle ABM is congruent to triangle PQN
(II)triangle ABC is congruent to triangle PQR
Two sides AB and BC and median AM of one triangle ABC are respectively equal to the sides PQ and QR and median PN of triangle PQR.
show that : (I) triangle ABM is congruent to triangle PQN
(II)triangle ABC is congruent to triangle PQR
A P
B M C Q N R
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)
Two sides AB and BC and median AM of one triangle ABC are respectively equal to the sides PQ and QR and median PN of triangle PQR.
show that : (I) triangle ABM is congruent to triangle PQN
(II)triangle ABC is congruent to triangle PQR
A P
B M C Q N R
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)
