Ratio Stories

Geometry Level 4

In the given diagram above, triangle A B C ABC and triangle E D C EDC are right angled at B B and D D respectively. A B = E D = 3 AB=ED=3 and B C = C D = 4 BC=CD=4 . B E BE is joined and line segment B E BE intersects A C AC at M M .

Find the ratio in which M M divides the line segment B E BE , i.e. find B M : M E BM:ME .

Note : Diagram not to scale.

1 : 2 1:2 3 : 7 3:7 11 : 19 11:19 1 : 3 1:3

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Arkajyoti Banerjee by Arkajyoti Banerjee , 20, India

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3 solutions

Construct MN ⊥ BC. Then, angle MNC = 90 degrees. Also, angle ABC = 90 degrees. Also, in tr. MNC and tr. ABC, angle MCN is common. Then tr. MNC is similar to tr. ABC. Then, Now in tr. BNM and tr. BDE, angle MBN is common and angle BDE and angle BNM both are 90 degrees. Therefore these two triangles are similar too. Then, From (i) and (ii), we get, Now by Thales’ theorem or by BPT (Basic Proportionality Theorem) in tr. BDE,

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Nice problem.

In step ( i ) (i) , it should be M N A B = B C − B N B C \frac{MN}{AB}= \frac{BC-BN}{BC}

Harsh Khatri - 5 years, 3 months ago

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Thanks. I've made the change.

Arkajyoti Banerjee - 5 years, 3 months ago
Surya Rathore
Mar 19, 2016

Draw AE . Hence, AE = BD = 8 and AE || BD

In Triangles EMA and BMC

Angle EMA = Angle BMC (Vertically Opposite Angles)

Angle MAE = Angle MCB (Alternate Interior Angles)

Angle MEA = Angle MBC (Alternate Interior Angles)

Therefore, Triangles EMA and BMC are congruent (AAA property of congruent triangles)

Now ME/AE = BM/CB

4x ME = 8x BM (AE = 8 and CB = 4)

ME = 2x BM

Therefore BM:ME = 1:2

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Moderator note:

Nice approach of recognizing the pairs of similar triangles that allow us to calculate the ratios.

EB is the diagonal of rectangle ABDE. Vertex A is connected to midpoint C of the opposite side, to cut the diagonal at M. So M trisects the diagonal BE. So the ratio BM : ME :: 1 : 2. No need of values 4 and 3.
O R . . . i n d e p e n d e n t p r o o f . OR ...independent ~~proof.
Let p= 1 2 ∗ B D \dfrac 1 2 *BD =BC=CD, q=NC. Drop MN perpendicular to BD. Therefore AB || MN || ED.

I n Δ E B D , M N A B = M N E D = B N B D = p − q 2 p . B N N D = B M M E I n Δ A B C , M N A B = N C B C = q p . ⟹ q p = p − q 2 p = p 3 p = 1 3 = B N B D . B N B D − B N = B N N D = 1 3 − 1 ⟹ B M M E = B N N D = 1 2 . In~~ \Delta~ EBD, ~ ~\color{#3D99F6}{\dfrac {MN}{AB}}=\dfrac {MN}{ED}=\color{#D61F06}{\dfrac{BN}{BD}}=\dfrac { p - q} {2p}.\\ \color{#E81990}{\dfrac{BN}{ND}=\dfrac{BM}{ME}}\\ In~~ \Delta~ ABC , ~~\color{#3D99F6}{\dfrac {MN}{AB}}=~\dfrac{NC}{BC}=\dfrac q p.\\ \implies ~\dfrac q p=\dfrac { p - q}{2p}=\dfrac p {3p} =\color{#D61F06}{\dfrac 1 3= \dfrac{BN}{BD} }.\\ \dfrac{BN}{BD - BN}=\dfrac{BN}{ND}=\dfrac 1 {3 - 1}~~ \implies~\color{#E81990}{\dfrac{BM}{ME}=\dfrac{BN}{ND}=\dfrac 1 2}.

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How can you conclude that MN/AB = BC/NC ?

Shouldn't it be MN/AB = NC/BC ?

Arkajyoti Banerjee - 5 years, 3 months ago

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You are right. Thank you for the correction.

Niranjan Khanderia - 5 years, 3 months ago

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