In Figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX DE meets BC at Y. Show that:
In Figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX DE meets BC at Y. Show that:
(i) ΔMBC ≅ ΔABD
(ii) ar(BYXD) = 2ar(MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2ar(FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN)+ar(ACFG)
Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
(i) We know that each angle of a square is 90°. Hence, ∠ABM = ∠DBC = 90º
∴ ∠ABM + ∠ABC = ∠DBC + ∠ABC
∴ ∠MBC = ∠ABD
In ∆MBC and ∆ABD,
∠MBC = ∠ABD (Proved above)
MB = AB (Sides of square ABMN)
BC = BD (Sides of square BCED)
∴ ∆MBC ≅ ∆ABD (SAS congruency)
(ii) We have
∆MBC ≅ ∆ABD
∴ ar (∆MBC) = ar (∆ABD) … (i)
It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)
∴ BD  AX (Two lines perpendicular to same line are parallel to each other)
∆ABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.
Area (∆YXD) = 2 Area (∆MBC) [From equation (i)] … (ii)
(iii) ∆MBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.
2 ar (∆MBC) = ar (ABMN)
ar (∆YXD) = ar (ABMN) [From equation (ii)] … (iii)
(iv) We know that each angle of a square is 90°.
∴ ∠FCA = ∠BCE = 90º
∴ ∠FCA + ∠ACB = ∠BCE + ∠ACB
∴ ∠FCB = ∠ACE
In ∆FCB and ∆ACE,
∠FCB = ∠ACE
FC = AC (Sides of square ACFG)
CB = CE (Sides of square BCED)
∆FCB ≅ ∆ACE (SAS congruency)
(v) AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC) [given]
Hence,
CE  AX (Two lines perpendicular to the same line are parallel to each other)
Consider BACE and parallelogram CYXE
BACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.
∴ar (∆YXE) = 2ar (∆ACE) … (iv)
We had proved that
∴ ∆FCB ≅ ∆ACE
ar (∆FCB) ≅ ar (∆ACE) … (v)
From equations (iv) and (v), we get
ar (CYXE) = 2 ar (∆FCB) … (vi)
(vi) Consider BFCB and parallelogram ACFG
BFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.
∴ ar (ACFG) = 2 ar (∆FCB)
∴ ar (ACFG) = ar (CYXE) [From equation (vi)] …....(vii)
(vii) From the figure, we can observe that
ar (∆CED) = ar (∆YXD) + ar (CYXE)
∴ ar (∆CED) = ar (ABMN) + ar (ACFG) [From equations (iii) and (vii)].

P and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the midpoint of AP, show that:
3 years ago

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD)×ar (BPC).
3 years ago

In Figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, show that:
3 years ago

In Figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).
3 years ago

In Figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
3 years ago
 321 Forums
 27.3 K Topics
 53.8 K Posts
 0 Online
 12.4 K Members