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Practice each question, then open the answer or worked solution.
ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that
BD = CE.
According to the question,
ΔABC is an isosceles triangle and AB = AC, BD and CE are two medians
From ΔABD and ΔACE,
AB = AC (given)
2 AE = 2 AD (as D and E are mid points)
So, AE = AD
∠A = ∠A (common)
Hence, ΔABD ≅ ΔACE (using SAS)
⇒ BD = CE (by CPCT)
Hence proved.
In Fig.7.4, D and E are points on side BC of a Δ ABC such that BD = CE and AD = AE.
Show that Δ ABD ≅ Δ ACE.
According to the question,
In ΔABC,
BD = CE and AD = AE.
In ΔADE,
AD = AE
Since opposite angles to equal sides are equal,
We have,
∠ADE = ∠AED … (1)
Now, ∠ADE + ∠ADB = 180° (linear pair)
∠ADB = 180° – ∠ADE … (2)
Also, ∠AED + ∠AEC = 180° (linear pair)
∠AEC = 180° – ∠AED
Since, ∠ADE = ∠AED
∠AEC = 180° – ∠ADE … (3)
From equation (2) and (3)
∠ADB = ∠AEC … (4)
Now, In ΔADB and ΔAEC,
AD = AE (given)
BD = EC (given)
∠ADB = ∠AEC (from (4)
Hence, ΔABD ≅ΔACE (by SAS)
CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that
Δ ADE ≅ Δ BCE.
According to the question,
CDE is an equilateral triangle formed on a side CD of a square ABCD.
In ΔADE and ΔBCE,
DE = CE (sides of equilateral triangle)
Now,
∠ADC = ∠BCD = 90°
And, ∠EDC = ∠ECD = 60°
Hence, ∠ADE = ∠ADC + ∠CDE = 90° + 60° = 150°
And ∠BCE = ∠BCD + ∠ECD = 90° + 60° = 150°
⇒ ∠ADE = ∠BCE
AD = BC (sides of square)
Hence, ΔADE ≅ΔBCE (by SAS)
In Fig.7.6, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that Δ ABC ≅ Δ DEF.
According to the question,
BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC.
In ΔABC and ΔDEF
BA = DE (given)
BF = EC (given)
∠A = ∠D (both 90°)
BC = BF + FC
EF = EC + FC = BF + FC (∵ EC = BF)
⇒ EF = BC
Hence, ΔABC ≅ ΔDEF (by RHS)
Q is a point on the side SR of a Δ PSR such that PQ = PR. Prove that PS > PQ.
Given: in ΔPSR, Q is a point on the side SR such that PQ = PR.
In ΔPRQ,
PR = PQ (given)
⇒ ∠PRQ = ∠PQR (opposite angles to equal sides are equal)
But ∠PQR > ∠PSR (exterior angle of a triangle is greater than each of opposite interior angle)
⇒ ∠PRQ > ∠PSR
⇒ PS > PR (opposite sides to greater angle is greater)
⇒ PS > PQ (as PR = PQ)