CAT 2025 Lesson : Lines & Triangles - Triangle & its Segments

In $\triangle$ ABC, points D and E are the mid-points of sides BC and CA respectively. The lines AD and BE intersect at point O. What is the ratio of the areas of quadrilateral CDOE and $\triangle$ ABC?

Solution

AD and BE are the medians of $\triangle$ ABC. In the figure below, we draw the third median CF, which will also pass through O, which is the centroid. The 3 medians divide the larger triangle into 6 equal triangles.

Let the area of each smaller triangle be 1 unit. ∴ Area of $\triangle$ ABC $=$ 6 units Area(CDOE) = Area($\triangle$ COD) + Area( $\triangle$ COE) = 2 units Ratio of areas of CDOE to $\triangle$ ABC = 2 : 6 = 1 : 3

Answer: $1 : 3$

O is the incentre of $\triangle$ ABC. The line joining A and O intersects BC at D. If $\angle$ADB = $100\degree$ and $\angle$OCD = $35\degree$, then ∠BOC = ?

Solution

The incentre is where the angle bisectors meet. Therefore, AO, BO and CO are angle bisectors of $\triangle$ ABC

$\angle$OCD $= \angle$OCA = $35\degree$ (CO is the $\angle$ bisector) $\angle$ADC $= 180\degree$ $-$ $\angle$ADB $= 80\degree$ (Linear Pair) In $\triangle$ ADC, $\angle$DAC + $80\degree$ + $70\degree$ $= 180\degree$ ⇒ $\angle$DAC $= 30\degree$ $\angle$BOC $= 90\degree$ + $\dfrac{1}{2}$ $\angle$BAC (Angle bisector Property) $= 90\degree$+ $\angle$DAC $= 120\degree$

Answer: $120\degree$

In $\triangle$ ABC, D lies on BC such that BD : DC = 1 : 3 and E lies on AC such that AE : EC = 2 : 3. If F lies on DE such that DF : FE = 1 : 2, then what is the ratio of areas of triangles ABC and AFE?

Solution

Let AH be perpendicular to BC. Therefore, AH is the altitude for the bases BC, BD and DC.

Area ( $\triangle$ ABD): Area ($\triangle$ ADC) $= \dfrac{1}{2} \times$ BD $\times$ AH : $\dfrac{1}{2} \times$ DC $\times$ AH $=$ BD : DC $=$ 1 : 3

Going forward, please note that when a line segment from a vertice to the side divides the side in the ratio of $a : b$, the two triangles so formed have areas in the ratio of $a : b$ as the altitudes to the divided sides remains the same..

Lets start with the area of the innermost triangles and use a variable $x$. Area ( $\triangle$ ADF): Area ($\triangle$ AFE) $=$ 1 : 2 ∴ Let the actual areas of $\triangle$ ADF and $\triangle$ AFE be $x$ and 2$x$ respectively. Area of $\triangle$ ADE $= x + 2x = 3x$ Area ($\triangle$ ADE) : Area ($\triangle$ CDE) $=$ 2 : 3 $\dfrac{3x}{ \mathrm{Area ( \triangle CDE)}} = \dfrac{2}{3}$ ⇒ $\mathrm{Area ( \triangle CDE)} = 4.5x$

Area of $\triangle$ ADC $= 3x + 4.5x = 7.5x$

Area ( $\triangle$ ABD) : Area ( $\triangle$ ADC) $=$ 1 : 3 ⇒ $\dfrac{ \mathrm{Area ( \triangle ABD)}}{7.5x} = \dfrac{1}{3}$ ⇒ Area ( $\triangle$ ABD) $= 2.5x$

∴ Area ( $\triangle$ ABC) $=$ Area ( $\triangle$ ABD) + Area ( $\triangle$ ADC) $= 2.5x + 7.5x = 10x$

Ratio of areas of triangles ABC and AFE $= 10x : 2x = 5 : 1$

Answer: $5 : 1$