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Geometry Problem: Finding EC in Triangle ABC
The problem involves a triangle ABC where a line segment DE is drawn parallel to BC, intersecting sides AB and AC at points D and E, respectively. We are given the lengths AD = 3, DB = 6, and AE = 4, and we need to find the length of EC.
Applying the Basic Proportionality Theorem
When a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides the two sides proportionally. This is known as the Basic Proportionality Theorem (BPT) or Thales's Theorem.
According to the BPT, for triangle ABC with DE || BC:
$ \frac{AD}{DB} = \frac{AE}{EC} $
Calculation Step-by-Step
- Substitute Given Values: Plug the known lengths into the BPT formula:
$ \frac{3}{6} = \frac{4}{EC} $
- Simplify the Ratio: Simplify the fraction on the left side:
$ \frac{1}{2} = \frac{4}{EC} $
- Solve for EC: Cross-multiply to find the value of EC:
$ 1 \times EC = 2 \times 4 $
$ EC = 8 $
Therefore, the length of EC is 8.
Similar Questions
- In an urban project, triangular plots with side ratios 3:4:5 are allocated. If the perimeter is 60 m, find the area.
- In $\Delta ABC \sim \Delta XYZ$ and $AB = 4 \text{ cm}, BC = 6 \text{ cm}, XY = 8 \text{ cm}$, what is the length of YZ?
- A right-angled triangle ABC has legs AB=6 cm and BC=8 cm. An altitude BD is drawn from the vertex B to the hypotenuse AC. What is the length of the altitude BD?
- If in $\Delta LMN$ and $\Delta OPQ$, $\angle L = \angle O$, $LM = OP$, and $\angle M = \angle P$, which rule proves $\Delta LMN \cong \Delta OPQ$?
- A triangle with all its angles less than $90^\circ$ has its Orthocenter located where?
- Triangle PQR has vertices P(1, 1), Q(4, 1), and R(1, 5). Triangle STU has vertices S(6, 2), T(9, 2), and U(6, 6). Are these two triangles congruent?
- Two triangles, $\Delta PQR$ and $\Delta STU$, have $PQ=ST$, $PR=SU$, and $\angle QPR = \angle TSU$. By what rule are they congruent?
- The Incenter of a triangle is located at $(3, 4)$. The length of the altitude from the Incenter to side AB is $2\text{ units}$. What is the radius of the triangle's incircle?
- If in triangle ABC, $DE \parallel BC$ and $AD/DB = \frac{1}{2}$, and $AE = 5\text{ cm}$, then $EC$ is:
- For a triangle with one interior angle greater than 90°, where is its Orthocenter located?
Important Questions from Triangles, Congruence and Similarity
The radius of the circumcircle of an equilateral triangle of √3 unit side, is:
If the ratio of the angles of a triangle is 3 : 5 : 7, find the value of the largest angle.
A. 36°
B. 60°
C. 84°
D. 15°
If ΔABC and Δ PQR are similar and \(\rm\frac{BC}{QR} = \frac{1}{3}\) , find \(\rm\frac{ar(\Delta PQR)} {ar(\Delta BCA)}\)
If the angles of a triangle are in the ratio of 2 : 5 : 8, then find the value of the smallest angle.
ABCD is a parallelogram. Side BC is produced to E such that BC = CE. Join AE which intersects side CD at P. The area of triangle ABE is:
